Mathematics - Quadratic Equation Question with Solution | TestHub

The given expression is a quadratic $f(x)=x^2-4x+7$. Since the coefficient of $x^2$ is $1$ (which is positive), the parabola opens upwards and has a minimum value. The minimum occurs at $x=-\frac{b}{2a}=-\frac{-4}{2(1)}=2$. Substituting $x=2$ into the expression gives the minimum value: $f(2)=(2{)}^2-4(2)+7=4-8+7=3$.